Factor
$$(x+5)^4+(x+3)^4-82$$

Question

Uniteasy Admin · Accepted Answer

Given $$(x+5)^4+(x+3)^4-82$$
Let
$$y = x+4$$n
then
$$(x+5)^4+(x+3)^4-82$$
$$=(y-1)^4+(y+1)^4-82$$
$$=(y^2-2y+1)^2+(y^2+2y+1)^2-82$$
$$=(y^2+1)^2-4y(y^2+1)+4y^2+(y^2+1)^2+4y(y^2+1)+4y^2-82$$
$$=2(y^2+1)^2+8y^2-82$$
$$=2y^4+12y^2-80$$
$$=2(y^4+6y^2-40)$$
$$=2(y^2-4)(y^2+10)$$
$$ =2(y-2)(y+2)(y^2+10) $$
Substitute (1) then
$$(x+5)^4+(x+3)^4-82= 2(x+6)(x+2)(x^2+8x+26)$$

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